Intro Questions

• Why are we creating simulated spatial processes?
• For Today:
  • Model Fitting

Simulating Spatial Process

Simulated Spatial Process: Exercise

How does the spatial process change with:

• another draw with same parameters?
• a different value of \(\phi\)
• a different value of \(\sigma^2\)
• adding a nugget term, \(\tau^2\)

Classical Model Fitting

• The classical approach to spatial prediction is rooted in minimizing the mean-squared error.
• This approach is often referred to as Kriging in honor of D.G. Krige a South African mining engineer.
• As a result of Krige’s work (along with others), point-level spatial analysis and geostatistical analysis are used interchangeably.

Mathematical Motivation

• Let \(\boldsymbol{Y} = \{Y(\boldsymbol{s_1}), \dots, Y(\boldsymbol{s_n}) \}\) be observations of a spatial process and \(n\) sites.
• Then \(Y(\boldsymbol{s_0})\) is a site where the spatial process has not been observed.
• The Goal: What is the best predictor for \(Y(\boldsymbol{s_0})\) given that \(\boldsymbol{Y} = \{Y(\boldsymbol{s_1}), \dots, Y(\boldsymbol{s_n}) \}\) was observed?

Visual Motivation

• The Goal: What is the best predictor for \(Y(\boldsymbol{s_0})\) given that \(\boldsymbol{Y} = \{Y(\boldsymbol{s_1}), \dots, Y(\boldsymbol{s_n}) \}\) was observed?
  

  source: airnow.gov

Mathematical Notation

• A linear predictor for \(Y(\boldsymbol{s_0})\), given \(\boldsymbol{Y}\) takes the form \(\sum_i l_i Y(\boldsymbol{s_i}) + \delta_0\)
• Using squared error loss, we’d seek to minimize \[E \left[Y(\boldsymbol{s_0}) - \left(\sum_i l_i Y(\boldsymbol{s_i}) + \delta_0\right) \right]^2\] as a function of \(l_i\) and \(\delta_0\).
• Describe and interpret \(l_i\) and \(\delta_0\)

Connection to variogram

• Recall the intrinsic stationarity assumption \[E[Y(\boldsymbol{s+h}) - Y(\boldsymbol{s})] = 0,\] thus \(\sum_i l_i = 1\) such that \[E[Y(\boldsymbol{s_0}) - \sum_i l_i Y(\boldsymbol{s_i})] = 0\]
• Following this logic, we would now minimize \[E [Y(\boldsymbol{s_0}) - \sum_i l_i Y(\boldsymbol{s_i}) ]^2 + \delta_0^2,\] thus \(\delta_0 = 0\).

Connection to variogram: part 2

• Define \(a_0 = 1\) and \(a_i = -l_i\), then we can rewrite \[E [Y(\boldsymbol{s_0}) - \sum_i l_i Y(\boldsymbol{s_i}) ]^2 \; \; \text{ as } \; \; E [\sum_{i=0}^n a_i Y(\boldsymbol{s_i}) ]^2\]
• It turns out that \[E [\sum_{i=0}^n a_i Y(\boldsymbol{s_i}) ]^2 = -\sum_i \sum_j a_i a_j \gamma(\boldsymbol{s_i} - \boldsymbol{s_j})\]
• In other words, minimizing the squared error, under assumptions, justifies the variogram.
• This is a contrained optimization of a quadratic form that is typically handled with a Lagrange multiplier. Khan Academy Refresher Video

Lagrange multipliers

To rewrite the constrained optimization in terms of \(l_i\) we get \[-\sum_{i=0}^n \sum_{j=0}^n a_i a_j \gamma(\boldsymbol{s_i} - \boldsymbol{s_j}) = -\sum_{i=1}^n \sum_{j=1}^n l_i l_j \gamma_{ij} + 2 \sum_{i=1}^n l_i \gamma_{0i},\] where \(\gamma_{ij} = \gamma(\boldsymbol{s}_i - \boldsymbol{s}_j)\) and hence \(\gamma_{0j} = \gamma(\boldsymbol{s}_0 - \boldsymbol{s}_j)\)

• Solving equations with this constraint requires partial derivatives and the use of Lagrange multipliers, typically denoted as \(\lambda\).

BLUP

• It turns out that the solution for the vector \(\boldsymbol{l}\) is \[\boldsymbol{l} = \Gamma^{-1} \left( \boldsymbol{\gamma_0} + \frac{(1 - \boldsymbol{1}^T \Gamma^{-1} \boldsymbol{\gamma_0})}{\boldsymbol{1}^T \Gamma^{-1} \boldsymbol{1}}\boldsymbol{1}\right),\] where \(\Gamma\) is an \(n \times n\) matrix with entries \(\Gamma_{ij} - \gamma_{ij}\) and \(\boldsymbol{\gamma_0}\) is the vector of \(\gamma_{0i}\) values.
• Then the Best Linear Unbiased Predictor is \(\boldsymbol{l}^T\boldsymbol{Y}\)
• This BLUP also requires an estimate of \(\gamma(\boldsymbol{h})\)

So what does this all mean …

Consider a small example on 1-dimension.

• Given the observed values, what should the predictions be at \(\boldsymbol{s}^{*}_1 = 4\) and \(\boldsymbol{s}^{*}_2 = 6\)

So what does this all mean …

What should the predictions be at \(\boldsymbol{s}^{*}_1 = 4\) and \(\boldsymbol{s}^{*}_2 = 6\)

Kriging Exercise:

• Recall \[\boldsymbol{l} = \Gamma^{-1} \left( \boldsymbol{\gamma_0} + \frac{(1 - \boldsymbol{1}^T \Gamma^{-1} \boldsymbol{\gamma_0})}{\boldsymbol{1}^T \Gamma^{-1} \boldsymbol{1}}\boldsymbol{1}\right),\]

• Define \(\gamma(h) = 1 - \exp(- \frac{h}{3})\) and compute the BLUPs for \(\boldsymbol{s}_1^{*}\) and \(\boldsymbol{s}_2^{*}\)

• Interpret and explain \(\boldsymbol{l}\) for each sample point.

• If you have time, fill in the line (rather than the surface) from (0.5, 7.5)

Kriging Solution

A Gaussian Process

Additional Resources

• Meuse Data Tutorial
• Textbook Data Sets